Math Problem Statement

When running a 100-meter race Ritchie reaches his maximum speed 3 seconds after the race started when he is 13.5 meters from the starting line. Ritchie runs at this max speed for the rest of the race. Ritchie is 48.5 meters from the starting line 8 seconds after the start of the race.

Solution

Let's break down the situation with the data provided:

  1. Phase 1: Acceleration phase (from 0 to 3 seconds)

    • Ritchie reaches his maximum speed 3 seconds after the race starts.
    • In this time, he covers 13.5 meters.
  2. Phase 2: Constant speed phase (from 3 seconds to the end of the race)

    • After 3 seconds, Ritchie maintains his maximum speed.
    • At 8 seconds after the start, Ritchie has covered 48.5 meters.

Step 1: Calculating the acceleration phase

The acceleration phase lasts 3 seconds, and in this time, Ritchie covers 13.5 meters. Assuming constant acceleration, we can use the equation for distance traveled under constant acceleration:

d=12at2d = \frac{1}{2} a t^2

Here:

  • d=13.5metersd = 13.5 \, \text{meters} (distance covered in 3 seconds)
  • t=3secondst = 3 \, \text{seconds}
  • aa is the acceleration.

Solving for aa:

13.5=12a(3)213.5 = \frac{1}{2} a (3)^2 13.5=12a(9)13.5 = \frac{1}{2} a (9) 13.5=4.5a13.5 = 4.5 a a=3m/s2a = 3 \, \text{m/s}^2

So, Ritchie's acceleration during the first 3 seconds is 3m/s23 \, \text{m/s}^2.

Step 2: Calculating Ritchie's maximum speed

Ritchie reaches his maximum speed at the end of the acceleration phase, which is after 3 seconds. We can calculate the maximum speed using the formula:

v=atv = a t

Where:

  • a=3m/s2a = 3 \, \text{m/s}^2
  • t=3secondst = 3 \, \text{seconds}

v=3×3=9m/sv = 3 \times 3 = 9 \, \text{m/s}

So, Ritchie's maximum speed is 9m/s9 \, \text{m/s}.

Step 3: Checking the distance at 8 seconds

Now we need to check how far Ritchie has traveled after 8 seconds.

  • In the first 3 seconds, he covers 13.5 meters.
  • In the next 5 seconds (from 3 to 8 seconds), he runs at his maximum speed of 9m/s9 \, \text{m/s}.

The distance covered in 5 seconds at constant speed is:

d=vt=9×5=45metersd = v t = 9 \times 5 = 45 \, \text{meters}

So, the total distance after 8 seconds is:

13.5+45=58.5meters13.5 + 45 = 58.5 \, \text{meters}

But the problem states that after 8 seconds, Ritchie has only covered 48.5 meters. There seems to be a discrepancy here. Would you like to adjust or clarify the information, or explore further possibilities, such as different interpretations of the data?


Let me know if you'd like more details on any part of the explanation or have any questions! Here are 5 related questions you might explore:

  1. What is the role of constant acceleration in determining speed?
  2. How can you derive time from distance and speed during constant motion?
  3. How do acceleration and speed interact during different phases of a race?
  4. Could deceleration ever be a factor in a sprint like this?
  5. How might environmental factors like wind or terrain affect Ritchie's speed?

Tip: When calculating distances during different phases of motion, always separate acceleration and constant speed phases for clarity.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Constant Acceleration
Speed
Distance-Time Relationship

Formulas

d = (1/2) a t^2
v = a t
d = v t

Theorems

Equations of Motion (under constant acceleration)

Suitable Grade Level

Grades 9-12